Question

Evaluate the double integral ∬dxyda where D is the triangular region with vertices (0,0),(5,0),(0,6).

Answer 1

To evaluate the double integral over the triangular region D, set up the integral with appropriate limits and solve the single-variable definite integral.

Step-by-step explanation:

To evaluate the double integral ∫∫ dxyda over the triangular region D with vertices (0,0), (5,0), and (0,6), we must first describe the limits of integration for this region. Since the triangle is right-angled, bounded by the x-axis, and the lines y=0 and x=5, we can express the upper bound of y in terms of x as y=6-(6/5)x. Remember that when setting up a double integral, the area element dA is typically represented as dx dy or dy dx depending on the order of integration.

In this case, we can integrate with respect to y first and then x. Our limits of integration for y will go from 0 to 6-(6/5)x, and for x from 0 to 5. We can therefore write the double integral as:

∫05∫06-(6/5)x dy dx. When we carry out the integration with respect to y, we simply get:

∫05[y]06-(6/5)x dx

After substituting the limits for y, our integral will then become:

∫05(6-(6/5)x) dx.

This is now a single-variable definite integral, and its evaluation will give us the area of the triangular region D.

Answer 2

To evaluate the double integral, we establish the limits for y as 0 to -6/5x and for x as 0 to 5, then integrate with respect to y first and then with respect to x to find the area of the triangular region D.

Step-by-step explanation:

To evaluate the double integral ∬ dxyda over the triangular region D with vertices (0,0),(5,0),(0,6), we first need to set up the integral limits by describing D in terms of x and y coordinates. Since the vertices suggest a right triangle with legs on the x and y axes, the limits of integration for y will go from 0 to a line that can be described by the equation y = mx + b, with m being the slope of the hypotenuse.

In this case, m = -6/5 because the hypotenuse runs from (0,6) to (5,0), and we can ignore the y-intercept b since it is 0. Therefore, the integration over y will go from 0 to -6/5x. The limits for x will simply be from 0 to 5.

The integral setup will be:

∬∬ dxyda = ∬ from 0 to 5 (∬ from 0 to -6/5x dy) dx

When evaluating the inner integral first with respect to y, we simply get y evaluated from 0 to -6/5x, resulting in -6/5x. Plugging this back into the outer integral gives us:

∬ from 0 to 5 (-6/5x dx)

By evaluating this integral, we find the area of the triangular region D. This is a fairly straightforward application of a double integral, where the order of integration can often be swapped depending on the simplicity of the functions involved in the region of integration.

The complete question is: Evaluate the double integral ∬dxyda where D is the triangular region with vertices (0,0),(5,0),(0,6). is: